Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $a = \dfrac{2k^2 - 6k}{k^2 - 8k + 15} \div \dfrac{5k^2 - 25k}{k^2 - 11k + 30} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{2k^2 - 6k}{k^2 - 8k + 15} \times \dfrac{k^2 - 11k + 30}{5k^2 - 25k} $ First factor out any common factors. $a = \dfrac{2k(k - 3)}{k^2 - 8k + 15} \times \dfrac{k^2 - 11k + 30}{5k(k - 5)} $ Then factor the quadratic expressions. $a = \dfrac {2k(k - 3)} {(k - 3)(k - 5)} \times \dfrac {(k - 5)(k - 6)} {5k(k - 5)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {2k(k - 3) \times (k - 5)(k - 6) } { (k - 3)(k - 5) \times 5k(k - 5)} $ $a = \dfrac {2k(k - 5)(k - 6)(k - 3)} {5k(k - 3)(k - 5)(k - 5)} $ Notice that $(k - 3)$ and $(k - 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {2k(k - 5)(k - 6)\cancel{(k - 3)}} {5k\cancel{(k - 3)}(k - 5)(k - 5)} $ We are dividing by $k - 3$ , so $k - 3 \neq 0$ Therefore, $k \neq 3$ $a = \dfrac {2k\cancel{(k - 5)}(k - 6)\cancel{(k - 3)}} {5k\cancel{(k - 3)}(k - 5)\cancel{(k - 5)}} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $a = \dfrac {2k(k - 6)} {5k(k - 5)} $ $ a = \dfrac{2(k - 6)}{5(k - 5)}; k \neq 3; k \neq 5 $